Question

An astronaut on a planet with no atmosphere drops a feather into a 6.87m deep crater and records that the feather falls freely for 1.32s.What is magnitude of free-fall acceleration on the planet? With what speed does the feather strike the bottom of the crater?

Answer 1

The vertical distance covered by freely falling feather is,

`h=ut+(1)/(2)at^2`

Since the feather is falling freely then the initial speed of feather is zero.

Plug in the known values,

`\begin{gathered} 6.87\text{ m=(0 m/s)t+}(1)/(2)a(1.32s)^2 \\ a=\frac{2(6.87\text{ m)}}{(1.32s)^2} \\ \approx\text{7}.89m/s^2 \end{gathered}`

Therefore, the free fall acceleration of the feather is 7.89 m/s2.

The speed of the feather at the bottom of the crater is,

`v=u+at`

Substitute the known values,

`\begin{gathered} v=0m/s+(7.89m/s^2)(1.32\text{ s)} \\ \approx10.4\text{ m/s} \end{gathered}`

Therefore, the speed of feather with which it strikes on the bottom of the crater is 10.4 m/s.